Proving that $mathbf{(H-frac{1}{n}J_n)}$ is indempotent
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I am trying to show that the matrix $mathbf{(H-frac{1}{n}J_n)}$ is idempotent where $mathbf{H}$ is the Hat-matrix (Projection matrix) of linear regression and $J_n$ is the $ntimes n$ matrix with $1$ in all its inputs. Taking :
$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)= HH - Hfrac{1}{n}J_n - frac{1}{n}J_nH + frac{1}{n}J_nfrac{1}{n}J_n}$$
Now, we know that $mathbf{H}$ and $mathbf{frac{1}{n}J_n}$ are idempontent, thus :
$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)=H-Hfrac{1}{n}J_n - frac{1}{n}J_nH +frac{1}{n}J_n}$$
How would I continue now in order to show that the given matrix is idempontent ?
linear-algebra matrices linear-regression idempotents
asked Nov 22 at 19:05
Rebellos
13.9k31243
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show 4 more comments
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1
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I am trying to show that the matrix $mathbf{(H-frac{1}{n}J_n)}$ is idempotent where $mathbf{H}$ is the Hat-matrix (Projection matrix) of linear regression and $J_n$ is the $ntimes n$ matrix with $1$ in all its inputs. Taking :
$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)= HH - Hfrac{1}{n}J_n - frac{1}{n}J_nH + frac{1}{n}J_nfrac{1}{n}J_n}$$
Now, we know that $mathbf{H}$ and $mathbf{frac{1}{n}J_n}$ are idempontent, thus :
$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)=H-Hfrac{1}{n}J_n - frac{1}{n}J_nH +frac{1}{n}J_n}$$
How would I continue now in order to show that the given matrix is idempontent ?
linear-algebra matrices linear-regression idempotents
asked Nov 22 at 19:05
Rebellos
13.9k31243
The "hat matrix" should be redefined, It is not a usual name...
– Jean Marie
Nov 22 at 19:08
@JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
– Rebellos
Nov 22 at 19:12
1
@JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
– Jean-Claude Arbaut
Nov 22 at 19:15
1
@Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
– Jean Marie
Nov 22 at 19:19
Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
– StubbornAtom
Nov 22 at 19:26
|
show 4 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to show that the matrix $mathbf{(H-frac{1}{n}J_n)}$ is idempotent where $mathbf{H}$ is the Hat-matrix (Projection matrix) of linear regression and $J_n$ is the $ntimes n$ matrix with $1$ in all its inputs. Taking :
$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)= HH - Hfrac{1}{n}J_n - frac{1}{n}J_nH + frac{1}{n}J_nfrac{1}{n}J_n}$$
Now, we know that $mathbf{H}$ and $mathbf{frac{1}{n}J_n}$ are idempontent, thus :
$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)=H-Hfrac{1}{n}J_n - frac{1}{n}J_nH +frac{1}{n}J_n}$$
How would I continue now in order to show that the given matrix is idempontent ?
linear-algebra matrices linear-regression idempotents
asked Nov 22 at 19:05
Rebellos
13.9k31243
I am trying to show that the matrix $mathbf{(H-frac{1}{n}J_n)}$ is idempotent where $mathbf{H}$ is the Hat-matrix (Projection matrix) of linear regression and $J_n$ is the $ntimes n$ matrix with $1$ in all its inputs. Taking :
$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)= HH - Hfrac{1}{n}J_n - frac{1}{n}J_nH + frac{1}{n}J_nfrac{1}{n}J_n}$$
Now, we know that $mathbf{H}$ and $mathbf{frac{1}{n}J_n}$ are idempontent, thus :
$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)=H-Hfrac{1}{n}J_n - frac{1}{n}J_nH +frac{1}{n}J_n}$$
How would I continue now in order to show that the given matrix is idempontent ?
linear-algebra matrices linear-regression idempotents
linear-algebra matrices linear-regression idempotents
asked Nov 22 at 19:05
Rebellos
13.9k31243
asked Nov 22 at 19:05
Rebellos
13.9k31243
edited Nov 22 at 19:13
asked Nov 22 at 19:05
Rebellos
13.9k31243
asked Nov 22 at 19:05
Rebellos
13.9k31243
asked Nov 22 at 19:05
Rebellos
13.9k31243
13.9k31243
The "hat matrix" should be redefined, It is not a usual name...
– Jean Marie
Nov 22 at 19:08
@JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
– Rebellos
Nov 22 at 19:12
1
@JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
– Jean-Claude Arbaut
Nov 22 at 19:15
1
@Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
– Jean Marie
Nov 22 at 19:19
Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
– StubbornAtom
Nov 22 at 19:26
|
show 4 more comments
The "hat matrix" should be redefined, It is not a usual name...
– Jean Marie
Nov 22 at 19:08
@JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
– Rebellos
Nov 22 at 19:12
1
@JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
– Jean-Claude Arbaut
Nov 22 at 19:15
1
@Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
– Jean Marie
Nov 22 at 19:19
Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
– StubbornAtom
Nov 22 at 19:26
The "hat matrix" should be redefined, It is not a usual name...
– Jean Marie
Nov 22 at 19:08
The "hat matrix" should be redefined, It is not a usual name...
– Jean Marie
Nov 22 at 19:08
@JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
– Rebellos
Nov 22 at 19:12
@JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
– Rebellos
Nov 22 at 19:12
1
1
@JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
– Jean-Claude Arbaut
Nov 22 at 19:15
@JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
– Jean-Claude Arbaut
Nov 22 at 19:15
1
1
@Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
– Jean Marie
Nov 22 at 19:19
@Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
– Jean Marie
Nov 22 at 19:19
Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
– StubbornAtom
Nov 22 at 19:26
Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
– StubbornAtom
Nov 22 at 19:26
|
show 4 more comments
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The "hat matrix" should be redefined, It is not a usual name...
– Jean Marie
Nov 22 at 19:08
@JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
– Rebellos
Nov 22 at 19:12
1
@JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
– Jean-Claude Arbaut
Nov 22 at 19:15
1
@Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
– Jean Marie
Nov 22 at 19:19
Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
– StubbornAtom
Nov 22 at 19:26